Question

(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

Answer 1

Answer with Explanation:w

a.We are given that

Potential difference, V=48 V

$R_1=24\Omega$

$R_2=96\Omega$

Equivalent resistance when R1 and R2 are connected in series

$R_{eq}=R_1+R_2$

Using the formula

$R_{eq}=24+96=120\Omega$

We know that

$I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A$

In series combination, current passing through each resistor is  same and potential difference across each resistor is different.

Power, P=$I^2 R$

Using the formula

Power,$P_1=I^2R_1=(0.4)^2\times 24=3.84 W$

Power, $P_2=I^2 R_2=(0.4)^2(96)=15.36 W$

b.

In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..

Current,$I=\frac{V}{R}$

Using the formula

$I_1=\frac{V}{R_1}=\frac{48}{24}=2 A$

$I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A$

$P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W$

$P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W$